Can you solve 5th question 4/1R₁ww+ V₁VinD₁D₂Figure 1: Circuit for Laboratory Experiment 34. Set Vin = 0.5 V and repeat the experiment.V₁=0,1709 VV₂V₂=0,171515. Comment on the on-voltages of D₁ and D₂.R₂/31 D₁ = 0,4127VD2=0,58311I = 0.1387 MA12=OI3 = 0,13702 D₁ = 0,5425VD2=016292✓3D₁In the circuit in Figure 1, D₁ is a germanium diode and D2 is a silicon diode. R₁ = R2 = 1.2 kn.1. Build the circuit on breadboard.2. Set Vin = 3V and measure the voltages on D₁ and D2. Measure the currents /1, 12 and 13as well as the voltages V₁ and Vz.IL = 11.5444 MA02=0₁17181V₁: 1199602015908€3= 014532 MA3. Set Vin = 5V and measure the voltages on D₁ and D2 again. Measure the remaining currents andvoltages.V₁= 3,8675 V₂ = 0,6303I1=2.9814= 0,1896V12= 2,2955 mAI3 = 0, 4829 MA6. Show that KVL and KCL hold for the circuit in Figure 1. 2 loop equations and 1 nodeequationare needed for KVL and KCL, respectively.Design a detailed report with the help of Report Writing Guide. Remember to use the notes youtook during the laboratory session. You are allowed to use the photographs you took during thesession. Do not forget to answer all questions.

given circuit is mentioned that D1 is germanium and D2 is silicon diode

Answered: 1=0,1409 V V₂=0,1715x 12=O 5. Comment…

Engineering

Electrical Engineering

1=0,1409 V V₂=0,1715x 12=O 5. Comment on the on-voltages of D₁ and D₂. 3 = 0,1370 ). 1387MA

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Concept explainers

Power Measurement

According to Blondel's theorem, the total power is measured with one less number of wattmeters than the number of wires present in the system. So, a single-phase and two-wire system will require one wattmeter and a single-phase and the three-wire system will require two wattmeters, and a three-phase and the three-wire system will also require two wattmeters and a three-phase and the four-wire system will require three wattmeter devices. The wattmeter is a device that measures power using one voltage and one current input as shown in the above figure. Multiple current or voltage input pair cables are used by many power analyzers and digital storage oscilloscopes (DSOs) for measuring watts, in effect within a single instrument acting as multiple wattmeters. It is possible to measure three-phase and four-wire power with one correctly specified power analyzer.

Question

Can you solve 5th question

4/1
R₁
ww
+ V₁
Vin
D₁
D₂
Figure 1: Circuit for Laboratory Experiment 3
4. Set Vin = 0.5 V and repeat the experiment.
V₁=0,1709 V
V₂
V₂=0,17151
5. Comment on the on-voltages of D₁ and D₂.
R₂
/3
1 D₁ = 0,4127V
D2=0,58311
I = 0.1387 MA
12=O
I3 = 0,1370
2 D₁ = 0,5425V
D2=016292✓
3
D₁
In the circuit in Figure 1, D₁ is a germanium diode and D2 is a silicon diode. R₁ = R2 = 1.2 kn.
1. Build the circuit on breadboard.
2. Set Vin = 3V and measure the voltages on D₁ and D2. Measure the currents /1, 12 and 13
as well as the voltages V₁ and Vz.
IL = 11.5444 MA
02=0₁17181
V₁: 1199602015908
€3= 014532 MA
3. Set Vin = 5V and measure the voltages on D₁ and D2 again. Measure the remaining currents and
voltages.
V₁= 3,8675 V₂ = 0,6303
I1=2.9814
= 0,1896V
12= 2,2955 mA
I3 = 0, 4829 MA
6. Show that KVL and KCL hold for the circuit in Figure 1. 2 loop equations and 1 nodeequation
are needed for KVL and KCL, respectively.
Design a detailed report with the help of Report Writing Guide. Remember to use the notes you
took during the laboratory session. You are allowed to use the photographs you took during the
session. Do not forget to answer all questions.

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